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secondary 3 | A Maths
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Can someone help me explain how to solve this problem? Thank you so much!
Date Posted: 4 years ago
Views: 262
(i) After a minute of use, the existing number of germs will be (0.79)N. After two minutes of use, the existing number of germs will be (0.79)(0.79)N= (0.79)^2 N . We can see that the number of germs present depends on the duration of usage. Thus, after n minutes, the existing number of germs will be (0.79)^n (N)
(ii) (0.79)^20 (N)= 0.008964N
x = (1- 0.008964) x 100 =99 (2 s.f.)
(iii) 0.008964 = e^20k
ln both side
-4.7145 = 20k
k= -0.236
Hopefully this helps.
(ii) (0.79)^20 (N)= 0.008964N
x = (1- 0.008964) x 100 =99 (2 s.f.)
(iii) 0.008964 = e^20k
ln both side
-4.7145 = 20k
k= -0.236
Hopefully this helps.
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